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To find partition function we can use series expansion. For Ising
model on the triangular lattice with transverse magnetic field:


in the limit $\frac J B \ll 1$

\begin{multline}
\ln(Z) = \ln(2 \cosh (\beta B)) + \\
+ 3 \ln (2 (\cosh(\beta \sqrt{4 B^2 + J^2}) \cosh(\beta J) )) - 6 \ln(2 \cosh
(\beta B))
\end{multline}

If we know $\ln(Z)$ we can find any thermodynamic variable.  To find Energy we
should use $E = - \frac {d} {d\beta} \ln(Z)$.

\begin{multline}
E = - 3 \frac{J \sinh \left(J \beta\right) + \sqrt{J^{2} + 4
B^{2}} \sinh \left(\beta \sqrt{J^{2} + 4 B^{2}}\right)}{\cosh\left(J
\beta\right) + \cosh \left(\beta \sqrt{J^{2} + 4 B^{2}}\right)} +
\\
 + 5 \frac{B \sinh\left(\beta B\right)}{\cosh\left(\beta B\right)}
\end{multline}

But if we want something that is more nice, we can expand our expression in the
limit $J / T \ll 1$. 

And we will get for $\ln(Z)$
\begin{equation}
\ln(Z) = \ln(2 \cosh(\beta B)) + \frac {3 J^2} {4 \cosh^2(\beta B) (\beta^2 + \frac {\beta} {2 B} \sinh(2
\beta B))}
\end{equation}

And for Energy we will get
\begin{multline}
E = - B \tanh\left(\beta B\right) + \\
- 0.75 \frac{\beta J^{2} \left(2 + \frac{\sinh\left(2 \beta B\right)}{2 \beta B}
+ \cosh\left(2 \beta B\right)\right)}{\cosh^{2}\left(\beta B\right)} + 1.5
\frac{B J^{2} \beta^{2} \left(1 + \frac{\sinh\left(2 \beta B\right)}{2 \beta B}\right) \tanh\left(\beta B\right)}{\cosh^{2}\left(\beta B\right)}
\end{multline}

I've compared this series with monte-carlo simulations.
Parameters of the simulation:
\begin{itemize}
	\item Hamiltonian $\hat{H} = -J \sum_{\langle i j
	\rangle}\hat{\sigma_i^z}\hat{\sigma_j^z} - B \sum_i \hat{\sigma_i^x}$, where
	$\sigma$ - Pauli matrices, 
  \item Size of the lattice - $16 \times 16 $
  \item $\beta$ discretization step - $\Delta \tau =0.05$
  \item $J = 1$
  \item Number of Monte Carlo steps $10000 \tau$, where $\tau$ is correlation
  time
  \item Number of Monte Carlo steps to reach equilibrium $500 \tau$, where $\tau$ is correlation
  time
  \item $\Delta \tau$ is fixed during all simulation. Change in number of layers
  will give change in temperature. Minimum number of layers for $B \neq 0$ is 2,
  so my maximum temperature is $\frac 1 {\Delta \tau L}|_{L = 2} = 10$.
  \item Lower temperature is $0.5$. If we increase number of layers we can reach
  lower temperature, but I've limited myself to 0.5
\end{itemize}

And now some pictures with Energy per site to compare different approaches:

\begin{figure}[h]
\center{\includegraphics[width=130mm]{../graphs/ferromagnetic_series_expansion/EB10.png}}
\caption{$B = 10$}
\end{figure}

\begin{figure}[h]
\center{\includegraphics[width=130mm]{../graphs/ferromagnetic_series_expansion/EB8.png}}
\caption{$B = 8$}
\end{figure}

\begin{figure}[h]
\center{\includegraphics[width=130mm]{../graphs/ferromagnetic_series_expansion/EB6.png}}
\caption{$B = 6$}
\end{figure}

\begin{figure}[h]
\center{\includegraphics[width=130mm]{../graphs/ferromagnetic_series_expansion/EB4.png}}
\caption{$B = 4$}
\end{figure}

\begin{figure}[h]
\center{\includegraphics[width=130mm]{../graphs/ferromagnetic_series_expansion/EB2.png}}
\caption{$B = 2$}
\end{figure}

\begin{figure}[h]
\center{\includegraphics[width=130mm]{../graphs/ferromagnetic_series_expansion/EB1.png}}
\caption{$B = 1$}
\end{figure}


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